Em tham khảo nha :
\(\begin{array}{l}
a)\\
2C{H_3}COOH + Mg \to {(C{H_3}COO)_2}Mg + {H_2}\\
b)\\
{n_{Mg}} = \dfrac{{4,8}}{{24}} = 0,2mol\\
{n_{C{H_3}COOH}} = 2{n_{Mg}} = 0,4mol\\
{m_{C{H_3}COOH}} = 0,4 \times 60 = 24g\\
{m_{ddC{H_3}COOH}} = \dfrac{{24 \times 100}}{{12}} = 200g\\
c)\\
{n_{{H_2}}} = {n_{Mg}} = 0,2mol\\
{V_{{H_2}}} = 0,2 \times 22,4 = 4,48l\\
d)\\
{n_{{{(C{H_3}COO)}_2}Mg}} = {n_{Mg}} = 0,2mol\\
{m_{{{(C{H_3}COO)}_2}Mg}} = 0,2 \times 142 = 28,4g\\
{m_{ddspu}} = 4,8 + 200 - 0,2 \times 2 = 204,4g\\
C{\% _{{{(C{H_3}COO)}_2}Mg}} = \dfrac{{28,4}}{{204,4}} \times 100\% = 13,89\%
\end{array}\)